Problem: Solve for $y$, $ -\dfrac{8}{3y - 2} = \dfrac{4y + 5}{12y - 8} + \dfrac{5}{3y - 2} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3y - 2$ $12y - 8$ and $3y - 2$ The common denominator is $12y - 8$ To get $12y - 8$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{8}{3y - 2} \times \dfrac{4}{4} = -\dfrac{32}{12y - 8} $ The denominator of the second term is already $12y - 8$ , so we don't need to change it. To get $12y - 8$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{5}{3y - 2} \times \dfrac{4}{4} = \dfrac{20}{12y - 8} $ This give us: $ -\dfrac{32}{12y - 8} = \dfrac{4y + 5}{12y - 8} + \dfrac{20}{12y - 8} $ If we multiply both sides of the equation by $12y - 8$ , we get: $ -32 = 4y + 5 + 20$ $ -32 = 4y + 25$ $ -57 = 4y $ $ y = -\dfrac{57}{4}$